Lanjutan 3 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 16.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{75}^{\circ },{285}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{75}^{\circ },{343}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{75}^{\circ },{344}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{75}^{\circ },{345}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{75}^{\circ },{346}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}}\\ & \cos (x-30{)}^{\circ }=\cos {45}^{\circ }\\ & \iff x-{30}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={75}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{15}^{\circ }(tm)\\ & k=1\Rightarrow x_1={75}^{\circ }+{360}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{15}^{\circ }+{360}^{\circ }={345}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {75}^{\circ },{345}^{\circ }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }<x<{360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{100}^{\circ },{330}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{30}^{\circ },{330}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{120}^{\circ },{300}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{60}^{\circ },{120}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{50}^{\circ },{300}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos x^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ & \cos x^{\circ }=\cos {30}^{\circ }\\ & \iff x=\pm {30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={30}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{30}^{\circ }(tm)\\ & k=1\Rightarrow x_1={30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{30}^{\circ }+{360}^{\circ }={330}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {30}^{\circ },{330}^{\circ }}\right\}\end{array}\end{array}$

Lanjutan 2 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 11.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{b}.{\displaystyle y=2\sin \left({\displaystyle \frac{1}{2}\pi -x}\right)}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x+{\displaystyle \frac{1}{6}\pi })}\\ \\ & \text{d}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi +x}\right)}\\ \\ & \text{e}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi -2x}\right)}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}2\pi ,\text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (2x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kirinya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-2\sin (3x+45{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=-2\sin (3x-15{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=-2\sin (3x-45{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=2\sin (3x+15{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=2\sin (3x-45{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kanan}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{3}={120}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin 3(x-k)\\ & \text{dengan}-k\text{adalah}\\ & \mathbf{\text{besar geseran ke kanan}}{15}^{\circ }\\ & \text{Jadi},y=2\sin 3(x-15{)}^{\circ }=2\sin (3x-45{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-\cos (2x-30{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-60{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=\cos (2x+30{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=\sin (2x-80{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=\sin (2x+60{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}1\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=\cos 2(x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{15}^{\circ }\\ & \text{Jadi},y=\cos 2(x+15{)}^{\circ }=\sin (2x+30{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$

Lanjutan Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{positif terkecil yang memenuhi}\\ & \sin x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{adalah}....}\\ \\ & \text{a}.{\displaystyle {30}^{\circ }}\\ \\ & \text{b}.{\displaystyle {60}^{\circ }}\\ \\ & \text{c}.{\displaystyle {120}^{\circ }}\\ \\ & \text{d}.{\displaystyle {240}^{\circ }}\\ \\ & \text{e}.{\displaystyle {300}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin x& =-\frac{1}{2}\sqrt{3}\\ \text{Gun}& \text{akan rumus persamaan}\\ & \text{sederhana, yaitu}:\\ \sin x& =-\sin {60}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })\\ & =\sin {240}^{\circ }\\ x& ={240}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}\cos x={\displaystyle \frac{2\sqrt{5}}{5}\text{maka nilai}}\\ & \cot x\left({\displaystyle \frac{\pi }{2}-x}\right)\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{2}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}}\\ \\ & \text{c}.{\displaystyle \frac{1}{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{7}}\\ \\ & \text{e}.{\displaystyle \frac{1}{8}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos x& =\frac{2\sqrt{5}}{5},\text{maka}\\ {\sin }^{2}x& +{\cos }^{2}x=1\\ \sin x& =1-{\cos }^{2}x\\ & =\sqrt{1-{\cos }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{2\sqrt{5}}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{20}{25}}}=\sqrt{{\displaystyle \frac{5}{25}}}={\displaystyle \frac{\sqrt{5}}{5}}\\ \cot & \left({\displaystyle \frac{\pi }{2}-x}\right)=\tan x,\text{maka}\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}\\ & ={\displaystyle \frac{\sqrt{5}}{2\sqrt{5}}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Periode dari fungsi}f(x)=-2\cos 3x\text{adalah}....\\ \\ & \text{a}.{90}^{\circ }\\ \\ & \text{b}.{100}^{\circ }\\ \\ & \text{c}.{120}^{\circ }\\ \\ & \text{d}.{150}^{\circ }\\ \\ & \text{e}.{180}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Periode dari}:f(x)=-2\cos 3x\\ & \text{adalah}\\ & ={\displaystyle \frac{{360}^{\circ }}{3}}\\ & ={120}^{\circ }\\ & \\ & \text{Ingat bahwa}\\ & f(x)=a\cos bx,\text{maka periodenya}\\ & ={\displaystyle \frac{{360}^{\circ }}{b}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Perhatikanlah grafik berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Gambar di atas adalah grafik fungsi dari}\\ & \text{a}.f(x)=\cos 2x\\ \\ & \text{b}.f(x)=\cos 3x\\ \\ & \text{c}.f(x)=3\cos x\\ \\ & \text{d}.f(x)=3\cos 3x\\ \\ & \text{e}.f(x)={\displaystyle \frac{1}{3}\cos x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Gambar cukup jelas}\\ & \text{dengan periode}{360}^{\circ }\\ & \text{gambar dari grafik}f(x)=3\cos x\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}{\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}=....}\\ \\ & \text{a}.4\\ \\ & \text{b}.2-\sqrt{2}\\ \\ & \text{c}.\sqrt{2}-2\\ \\ & \text{d}.-4\\ \\ & \text{e}.-1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}}\\ & ={\displaystyle \frac{\sin {150}^{\circ }+\tan {180}^{\circ }+\cos {180}^{\circ }}{\sin {\displaystyle {90}^{\circ }+\cos {360}^{\circ }-3\cos {\displaystyle {60}^{\circ }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}+0+(-1)}}{1+1-3\left({\displaystyle \frac{1}{2}}\right)}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}}}{2-{\displaystyle \frac{3}{2}}}}\\ & =-{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =-1\end{array}\end{array}$

Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 1.& \text{Nilai}{75}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 75\times 1^{\circ }& =75\times {\displaystyle \frac{\pi }{180}radian}\\ {75}^{\circ }& ={\displaystyle \frac{5}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\tan \theta ={\displaystyle \frac{5}{12}\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{maka}\cos \theta \text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{5}{13}}\\ \\ & \text{b}.{\displaystyle \frac{12}{13}}\\ \\ & \text{c}.{\displaystyle \frac{13}{5}}\\ \\ & \text{d}.{\displaystyle \frac{13}{12}}\\ \\ & \text{e}.{\displaystyle \frac{12}{5}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah gambar segitiga berikut}\end{array}$.

$.\begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan \theta & ={\displaystyle \frac{5}{12},\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{lihat gambar di atas}\\ & \text{dengan dalil Pythagoras akan}\\ & \text{didapatkan sisimiringnya}=13\\ \text{jadi}& ,\text{nilai dari}\\ \cos \theta & ={\displaystyle \frac{12}{13}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Panjang BC adalah}....\\ & \text{a}.{\displaystyle 20\sin {36}^{\circ }}\\ & \text{b}.{\displaystyle 20\cos {36}^{\circ }}\\ & \text{c}.{\displaystyle 20\tan {36}^{\circ }}\\ & \text{d}.{\displaystyle 15}\\ & \text{e}.{\displaystyle 16}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan {36}^{\circ }& ={\displaystyle \frac{BC}{20}}\\ \iff & BC=20\tan {36}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\tan {300}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {300}^{\circ }& =\tan ({360}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}\tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{6}\sqrt{6}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{2}\sqrt{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\\ & =\tan {60}^{\circ }-\sin ({180}^{\circ }-{60}^{\circ })-\tan ({180}^{\circ }+{30}^{\circ })\\ & =\tan {60}^{\circ }-\sin {60}^{\circ }-\tan {30}^{\circ }\\ & =\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{3}\sqrt{3}}}\\ & =(1-{\displaystyle \frac{1}{2}-\frac{1}{3}})\sqrt{3}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}\end{array}$

Grafik Fungsi Trigonometri

Sebelumnya telah diketahui perbandingan trigonometri diberbagai kuadan dan sudut-sudut yang berelasi, selanjutnya dapat digambarkan garfik fungsinya, yaitu : $y=\sin x$, $y=\cos x$, dan $y=\tan x$.

A. Grafik Fungsi Sinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\sin b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

B. Grafik Fungsi Cosinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\cos b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

C. Grafik Fungsi Tangen

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\tan b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{180}^{\circ }}{b}\text{atau}{\displaystyle \frac{\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{nilai minimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& ={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{\displaystyle \frac{\pi }{6}})\text{atau boleh juga}}\\ & \text{dituliskan dengan bentuk}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{30}^{\circ })}\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|{\displaystyle \frac{4}{5}}\right|=\frac{4}{5}\\ \text{c}.& \text{Nilai minimumnya}:-\left|{\displaystyle \frac{4}{5}}\right|=-\frac{4}{5}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:{30}^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi}f(x)=2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & =2\cos 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =2\cos 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|2\right|=2\\ \text{c}.& \text{Nilai minimumnya}:-\left|2\right|=-2\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui fungsi}f(x)=\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & =\tan 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =\tan 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{180}^{\circ }}{2}}\right|={90}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\mathit{\text{tidak ada}}\\ \text{c}.& \text{Nilai minimumnya}:\mathit{\text{tidak ada}}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.